This iterative method reveals an answer of 100 mg as the mnimum
residual level.

An algebraic method considers that the minimum level is yielded
again after the addition of a 100 mg dose after a period of 1 half
life. So:

$\frac{(k + 100)}{2} = k$
$k + 100 = 2k$
$\therefore k = 100$

The half life, $\text t_\frac{1}{2}$= 6 days at the longest for an
individual patient. If a dose of 20mg is given every day and a
decay of $(\frac{1}{2})^\frac{1}{6}$is experience on a single day then the minimum level
is given by a formula similar to those previously derived:

$(k + 20)\times(\frac{1}{2})^\frac{1}{6} = k$

$k + 20 = k (\frac{1}{2})^\frac{-1}{6}$

$k(1- (\frac{1}{2})^\frac{-1}{6}) = -20$

$k = \frac{20}{(\frac{1}{2})^\frac{-1}{6} -1}$

$k = 163.3$ mg (1dp)

Therefore the peak value is obtained by adding 20 mg as this is the
dose given once this minimum level is obtained at the start of
every day. This gives a value of 183.3 mg (1dp) as a peak level
for an individual where it is the case that the drug has the
longest possible half life.

If the calculation is repeated for t$_\frac{1}{2} = 4$ days, then
the calculation above only requires a simple modification as a
decay of $(\frac{1}{2})^{1}{4}$ is now experienced on a daily
basis.

$(k + 20)\times(\frac{1}{2})^\frac{1}{4} = k$

$k + 20 = k (\frac{1}{2})^\frac{-1}{4}$

$k(1- (\frac{1}{2})^\frac{-1}{4}) = -20$

$k = \frac{20}{(\frac{1}{2})^\frac{-1}{4} -1}$

$k = 105.7$ mg (1dp) which gives a peak level of 125.7 mg (1dp).

An iterative method for the above problem may also be used. For
example:

$20\times(\frac{1}{2})^\frac{1}{6} \rightarrow 17.82$

$(17.82 + 20)\times(\frac{1}{2})^\frac{1}{6}\rightarrow
33.69$

...

$(Ans + 20)\times(\frac{1}{2})^\frac{1}{6}$

Repeatedly computing this into a calculator will eventually yield
the value of 163.3 mg as the minimum level. A smimlar approach may
be utilised for the t$_\frac{1}{2} = 4$ days case .

This next part of the question is a little bit trickier, but
consideration of relationships in a similar format to those already
discussed provide a neat solution.

First consider the case where t$_\frac{1}{2} = 6$ days. We are
looking for a situation where a dose given to a minimum level at
the end of the week results in a peak value of 183.3 mg. If
$\alpha$ = minimum residual drug level and $k$= weekly dose,
then:

$\alpha + k = 183.3$

and

$(\alpha + k) (\frac{1}{2})^\frac{7}{6} = \alpha$

as the effective number of half lives during the week is
$\frac{7}{6}$ and the residual level $\alpha$ must be obtained at
the end of the week. Substitution of the the first form into the
second gives the value of $\alpha$:

$183.3 \times (\frac{1}{2})^\frac{7}{6} = \alpha$

$\alpha = 81.6$ mg

$k = 183.3 - 81.6$

$k= 101.7$ mg (1dp)

for the case where t$_\frac{1}{2} = 4$, the calculation
becomes:

$\alpha + k = 125.7$

$(\alpha + k) (\frac{1}{2})^\frac{7}{4} = \alpha$

$\alpha = 37.4$ mg

$k = 125.7 - 37.4$

$k = 88.3$ mg (1dp)

Any equivalent iterative methods are also valid, though an
algebraic solution proves much easier to handle at this
stage.

So in summary the lowest and highest long term levels of drug in
the body are:

The calculations additionally be attempted using t$_\frac{1}{2} =
5$ days which is an average of the two values given, though the
insight given by using the upper and lower values of the half life
is probably more insightful.

This raises several issues for a patient. A drug is a
physiologically active substance and as such needs to be present in
levels between doses that are sufficient to cause an effect.
However drug levels in the body that are very high can cause
unwanted side effects to become more pronounced. Add this to the
fact that the half life of a drug in a patient can be highly
variable and you can see that the situation is quite complex!
Fluxoxetine is often marketed as Prozac and is an antidepressant
approved for the treatment of major depression, obsessive
compulsive disorder, bulimia and many other conditions. This
example illustrates a case: drug level stability is an essential
consideration as the underlying disorders the drug treats are
highly serious and concern some kind of mental instability.

Thus if a patient has a weekly dose then if the half life of the
drug in the body is 6 days then the peak and minimum levels
calculated may be sensible depending on those required for the drug
to be active. However, if the half life is 4 days then the long
term minimal level may be too low! Consideration of daily dosing
may mean that this option of drug delivery is not favourable for
individuals where the half life of the drug is 6 days as
persistently high levels of the drug may cause more manifest side
effects to be expressed. This question is more open to intuitive
interpretation of the data.

Drug companies would have to find a dosing level that would give a
suitable peak drug level for the majority of their patients, taking
into account the variability in half lives. As an interesting
aside, consider that a drug company decides that dosing should
occur every 3 days - will the patient remember when to take the
drug for a period of a month? Companies often resort to producing
packets of pills, a proportion of which don't
contain the physiologically active compound to ensure
the patient takes the pills regularly and obtains the correct
dose.